let n, m be Element of NAT ; :: thesis: ( n <= m & m <= n + 3 & not m = n & not m = n + 1 & not m = n + 2 implies m = n + 3 )
assume that
A1: n <= m and
A2: m <= n + 3 ; :: thesis: ( m = n or m = n + 1 or m = n + 2 or m = n + 3 )

per cases ( m <= n + 2 or m > n + 2 ) ;

suppose m <= n + 2 ; :: thesis: ( m = n or m = n + 1 or m = n + 2 or m = n + 3 )

hence ( m = n or m = n + 1 or m = n + 2 or m = n + 3 ) by A1, Th2; :: thesis: verum

end;

suppose m > n + 2 ; :: thesis: ( m = n or m = n + 1 or m = n + 2 or m = n + 3 )

then ( m = n + 2 or m = (n + 2) + 1 ) by A2, NAT_1:9;
hence ( m = n or m = n + 1 or m = n + 2 or m = n + 3 ) ; :: thesis: verum

end;

end;